The $n\text{th}$ partial sum of the series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ is given by ${{S}_{n}}=3-\frac{8}{n}$. $a_8=$
Solution: ${{a}_{8}}={{S}_{8}}-{{S}_{7}}$ ${{a}_{8}}=\left(3-\frac{8}{8}\right)-\left(3-\frac{8}{7}\right)=-1+\frac{8}{7}=\frac{1}{7}$